Edit 2: It should be correct now, let me know what you think
Major edit (feel free to read anyway and share your comments): Right when I posted this, I realized that $f$ should be fixed, whereas in my example with the brownian motion I use many different realizations of brownian motion over $[0,T]$.
Consider a closed interval $[0,T]$ and a partition $\sigma=\{0=t_0<t_1<\dots<t_m=T\}$. A function $f:[0,T]\rightarrow\Bbb R$ is said to have bounded variation if there exists a $M\in\Bbb R$ such that for any partition $\sigma$ we have $\sum\limits_{j=1}^m|f(t_{j-1})-f(t_j)|\le M$.
This definition of bounded variation is equivalent to $\sup\limits_\sigma \sum\limits_{j=1}^m|f(t_{j-1})-f(t_j)|<\infty$
where $\sup$ is taken over all partitions (of any size $m+1\in\Bbb N^*$).
What if we replaced the definition with
$f$ has finite variation if $\forall \sigma\exists M\in\Bbb R$ such that $\sum\limits_{j=1}^m|f(t_{j-1})-f(t_j)|\le M$
Now the bound $M$ is dependent on $\sigma$ and it doesn't seem so obvious that we can replace the definition with the statement about the $\sup$. Because in general we could have a set of finite objects with infinite supremum (for example the sequence $(n)_{n\ge1}$ has bounded terms for every $n$ but diverges to infinity).
I want to say that brownian motion is continuous so for every partition (no matter how small the smallest interval $[t_i,t_{i+1}]$ is) we will have $|B_{t_i}-B_{t_{i-1}}|$ is finite for every $i$ and so the variation over any $\sigma$ is at most $m$ times the largest difference $|B_{t_j}-B_{t_{j-1}}|$ say, but still finite. But I feel like taking the supremum over all $\sigma$ will produce infinity.
Maybe a probabilistic argument would work, reasoning with the expected value of $|B_{t_i}-B_{t_{i-1}}|$.
The increment $B_{t_i}-B_{t_{i-1}}$ follows a normal distribution with variance $t_i-t_{i-1}>0$. So $$\Bbb E[|B_{t_i}-B_{t_{i-1}}|]=\int\limits_{-\infty}^\infty|x|{1\over \sqrt{2\pi(t_i-t_{i-1})}}\exp(\frac{-x^2}{ 2 {(t_i-t_{i-1})}})dx=\sqrt{2(t_i-t_{i-1})\over\pi}$$
By linearity of the expectation $\Bbb E[\sum\limits_{i=1}^m|B_{t_i}-B_{t_{i-1}}|]=\sum\limits_{i=1}^m\Bbb E[|B_{t_i}-B_{t_{i-1}}|]=\sum\limits_{i=1}^m\sqrt{2(t_i-t_{i-1})\over\pi}\le \sqrt{\frac{2m}{\pi}T}$
where in the last inequality I use Jensen's inequality with concave $x\mapsto\sqrt{x}$ and the fact that $\sum\limits_{i=1}^mt_i-t_{i-1}=T-0=T$
This is not very useful, we need a lower bound. Suppose $m$ is large enough (and we want to prove a result as $m\rightarrow \infty$) so that for every $i$ we have $t_i-t_{i-1}<1$. This implies that $\sqrt{t_i-t_{i-1}}>t_i-t_{i-1}$.
So we get $$\Bbb E[\sum\limits_{i=1}^m|B_{t_i}-B_{t_{i-1}}|]>T\sqrt{\frac{2}{\pi}}$$
This is better but still not good enough. Choose $\sigma$ so that mesh($\sigma$):=$\max\limits_{i\in\{1,\dots,m\}}|t_i-t_{i-1}|\le\frac{T}{m}$. In fact if we divide $[0,T]$ into $m$ intervals and mesh$(\sigma)\le\frac{T}{m}$ then they have to be equally spaced so $t_i-t_{i-1}=\frac{T}{m}\ \forall i$. Then we get $$\Bbb E[\sum\limits_{i=1}^m|B_{t_i}-B_{t_{i-1}}|]=\sqrt{2\over\pi}\sum\limits_{i=1}^m\sqrt{T\over m}=\sqrt{2\over\pi}m\sqrt{T\over m}=\sqrt{{2m\over\pi}T}\xrightarrow[m\rightarrow\infty]{\infty}$$
This proves that for a fixed realization of brownian motion on $[0,T]$ one can find a sequence of partitions $(\sigma_m)_{m\in\Bbb N^*}$ of $[0,T]$, namely the equally spaced partitions of size $m$, for $m\in\Bbb N^*$, such that the variation of this BM is equal to $\sqrt{{2m\over\pi}T}$ in expectation.
By reasoning probabilistically (meaning that there are possible realizations of a random variable that fall higher than its mean), for every $M\in \Bbb R$ there exists a partition $\sigma=\{0=t_0<t_1<\dots<t_k=T\}$ (where we can choose $k\ge\frac{(M+1)^2\pi}{2T} $) with an expected variation of at least $M+1$. Here it is important to point out that there a given partition might produce a variation that is smaller than $\sqrt{{2m\over\pi}T}$ but the important thing is that there are infinitely many $m\in\Bbb N^*$ for which the variation falls around (or higher than) $\sqrt{{2m\over\pi}T}$. This is the part where I'm a bit fuzzy.
Is this a proof of the following statement?
A realization Brownian Motion has the property of having finite variation for any finite partition but $\sup\limits_\sigma \sum\limits_{i=1}^m|B_{t_i}-B_{t_{i-1}}|=\infty$
I would love to get some feedback on that, I wasn't planning to write a complete proof like that, got a bit carried away with the "do your own research" rule.